On the Tight Chromatic Bounds for a Class of Graphs without

On the Tight Chromatic Bounds for a Class of Graphs without Three Induced Subgraphs Medha Dhurandhar

Abstract: Here we prove that a graph without some three induced subgraphs has chromatic number at the most equal to its maximum clique size plus one.

Introduction: It has been an eminent unsolved problem in graph theory to determine the chromatic number of a given graph. Failing in the efforts to determine this, attempts have been made to find good bounds for the chromatic number of a graph. Vizing [4] proved that if a graph does not induce some nine subgraphs, then (G)  (G)  (G)+1 where  (G) is the size of maximum clique in G and  (G) is the chromatic number of G. Later Choudum [1] and Javdekar [2] improved this result by dropping five and six of these nine subgraphs from the hypothesis, respectively. Finally Kierstead [3] showed that (G)  (G)  (G)+1 for a {K1,3,(K5-e)}-free graph. Furthermore, Dhurandhar [5] proved that (G)  (G)  (G)+1 for a {K1,3, (K2K1)+K2}-free graph. In this paper we prove that if G is {K1,3, H1, H2}-free, then (G)  (G)+1 where

H1 =

, and H2 =

Notation: For a graph G, V(G), E(G), (G), (G), (G) denote the vertex set, edge set, maximum degree, size of a maximum clique, chromatic number respectively. For u  V(G), N(u) = {v  V(G) / uv  E(G)}, and N (u ) = N(u)(u). If S  V(G), then denotes the subgraph of G induced by S. Also for u  V(G), deg u and denote the degree of u in G and a maximum clique in N(u). If C is some coloring of G and if a vertex u of G is colored m in C, then u is called a m-vertex. All graphs considered henceforth are simple. First we prove a Lemma, which will be used later. Lemma 1: Let G be {K1,3, H1, H2}-free and v  V(G). Then either 1. is complete or 2. {N(v)-Qv} = {w, z}, zw  E(G). Further  w’, z’  Qv s.t. ww’, zz’  E(G) and zq  E(G)  q  Qv-z’, wq  E(G)  q  Qv-w’ or 3. {N(v)-Qv} = {w, z, x} with xz, xw  E(G) and wz  E(G). Further  w’, z’  Qv s.t. ww’, zz’, xw’, xz’  E(G) and wq  E(G)  q  Qv-w’, zq  E(G)  q  Qv-z’, xq  E(G)  q  Qv{w’, z’}. Proof: Let  v  V(G) with w, z  N(v)-Qv s.t. wz  E(G). Then  w’, z’  Qv s.t. ww’, zz’  E(G). As G is K1,3-free, wz’, zw’  E(G) and  q Qv-{w’, z’}, qw  E(G) or qz  E(G). W.l.g. let qw  E(G). Then qz  E(G) (else = H1). This proves 2, in case |N(v)-Qv| = 2. Let |N(v)-Qv|

> 2 and {N(v)-Qv}  {w, z, x} with wz  E(G). As before  w’, z’  Qv s.t. ww’, zz’  E(G) and wq  E(G)  q  Qv-w’, zq  E(G)  q  Qv-z’. W.l.g. let xw  E(G). Claim: xz  E(G) If not, then xz’  E(G) and as   3,  q  Qv-{w’, z’}. Now qz  E(G) iff qw  E(G) (else = H1). Hence as G is K1,3-free, qw, qz  E(G). Again xq  E(G) (else = H1). But then is a bigger clique in N(V) than Qv, a contradiction. This proves the Claim. Again xq  E(G)  q  Qv-{w’, z’} (else xw’  E(G) and = H2) and hence xw’, xz’  E(G). Now if |N(v)-Qv| > 3, then let y  N(v)-Qv. As before yq  E(G)  q  Qv-{w’, z’} and yw’, yz’  E(G). But then if xy  E(G) is a bigger clique in N(V) than Qv and if xy  E(G) = K1,3, a contradiction. Hence |N(v)-Qv| = 3 and this proves 3.

This completes the proof of Lemma 1. For completeness we simply state a result from [6], which will be used in the main result. Lemma 2: If G is K1,3-free, then each component of the subgraph of G induced by two colour classes is either a path or a cycle. Theorem: If G is a {K1,3, H1, H2}-free, then (G)  (G)+1. Proof: Let G be a smallest {K1,3, H1, H2}-free graph with (G) > (G)+1. Now by minimality  u  V(G), (G-u)  (G-u)+1. Thus (G)+1 < (G)  (G-u)+1  (G-u)+2  (G)+2. Hence  u  V(G), (G-u) = (G)+1 and hence (G) = (G)+2. Case 1:  v  V(G) s.t. {N(v)-Qv} = {w, z} with zw  E(G). Let w’, z’  Qv be s.t. ww’, zz’  E(G). Then by Lemma 1, zq  E(G)  q  Qv-z’, wq  E(G)  q  Qv-w’. Let C =

 2

 i be a ((G)+2)-coloring of G in which v receives the unique color +2 and 1

vertices in Qv receive colors 1, ..., |Qv|. W.l.g. let w be the -vertex and z be the (+1)-vertex of v. Then by Lemma 2,  a w-z path P s.t. vertices on P are alternately colored  and +1 (else w can be colored by +1 and v by ). Let x be the +1-vertex adjacent to w on P. Then xz’  E(G) (else = H1 or H2). But then = H1, a contradiction. Case 2:  v  V(G) s.t. {N(v)-Qv} = {w, z, x} with xz, xw  E(G) and wz  E(G). Let w’, z’  Qv be s.t. ww’, zz’  E(G). Then by Lemma 1, xw’, xz’  E(G), wq  E(G)  q  Qvw’, zq  E(G)  q  Qv-z’, and xq  E(G)  q  Qv-{w’, z’}. Let C =

 2

 i be a ((G)+2)-coloring of 1

G in which v receives the unique color +2 and vertices in Qv receive colors 1, ..., |Qv|. Clearly colors  and +1 are used in N(v)-Qv (else color v by the missing color). Case 2.1: x has color say . Again N(v)-Qv has a +1-vertex (else color v by +1). Thus either w’ or z’ has a color not used in N(v)-Qv. W.l.g. let z’ have a color not used in N(v)-Qv. As z’x  E(G), z’ has a -vertex say y outside N(v). Then yw  E(G) (else = H1). But then = H1 or H2, a contradiction. Case 2.2: x has color other than  and +1. Then w.l.g. let w, z be colored by  and +1 resply. As before either w’ or z’ has a color not used in N(v)-Qv. W.l.g. let z’ have a color not used in N(v)-Qv. As z’z  E(G), z’ has a +1-vertex say y outside N(v). Then yw  E(G) (else = H1 or H2). But then = H1, a contradiction. Case 3:  v  V(G), is complete Let Q be a maximum clique in G and v  Q. Let C =

 2

 i be a ((G)+2)-coloring of G in which v 1

receives the unique color +2 and vertices in Q receive colors 1, ..., -1. Label v as v0. Clearly colors  and +1 are used in N(v)-Q (else color v by the missing color). Hence as |N(v)-Q|  |Q|,  a color say 1 not used in N(v)-Q. Label the vertex in Q, which has color 1 as v1 ( such a vertex, else color v by 1). Again  a color say 2 not used in N(v1)-Q and a 2-vertex in Q (else color v1 by 2 and v by 1). Label this 2-vertex in Q as v2 and so on. Label v as v0. Let v0, v1,..., vk be a maximal sequence of vertices in Q s.t. vi has color i and color i+1 is not used in N(vi)-Q, 1ik. By maximality of the sequence k+1= t+1 for some 0tk-2. Consider a component P containing -vertex of vt s.t. vertices in P are colored t+1 or . As vt has a unique -vertex, by Lemma 2, P is a path. If vt+1  P, then alter colors of vertices in P, color vt by  and vj by j+1 for 0jt-1, a contradiction. Hence vt+1  P. Similarly if R is a component containing -vertex of vk s.t. vertices in R are colored t+1 or , then vt+1  R. Thus P=R and as G is K1,3-free at least two of vt, vt+1, vk have the same -vertex. Similarly at least two of vt, vt+1, vk have the same +1-vertex. Case 3.1: At the most two of vt, vt+1, vk have the same -vertex. Let {m, n, p} = {t, t+1, k} and vm, vn have the same -vertex say x and vp have the -vertex say y. Then < vm, vn, x, vp, y> = H1, a contradiction. Case 3.2: All three have the same -vertex say x and the same +1-vertex say y. Now xy  E(G) (else x has another +1-vertex say z and = H1). W.l.g. let xvi  E(G). Then vi has no -vertex (else if z is the -vertex of vi, then = H1). Thus i > k. Again each of vt, vt+1, vk has another i-vertex (else color vi by , vj by i and vl by l+1 for l < j where j  {t, t+1, k}). Let z be an i-vertex s.t. z  vi and zvt  E(G). As G is K1,3-free, xz  E(G) and x has at the most two i-vertices. Hence at least two of vt, vt+1, vk have the same i-vertex other than vi. If {m, n, p} = {t, t+1, k} and say vm has a different i-vertex say w, then = H1>, a contradiction. Thus all of vt, vt+1, vk have the same i-vertex say z and xz  E(G). As is a maximum clique, clearly  i, j s.t. xvi, xvj  E(G) and zw  E(G) where w is the other j-vertex of vt, vt+1, vk. Clearly zvj, wvi  E(G). Now if yvi  E(G), then as before yz  E(G). But then if yvj  E(G), yz  E(G) and = H1 and if yvj  E(G), then = H1, a contradiction. Hence yvj, yvi  E(G). Clearly  s s.t. yvs E(G). As before xvs  E(G). But then = H2, a contradiction.

This proves the theorem. Examples to show that the Upper Bound is Tight: Let G = C2n+1, n > 1. Then G is {K1,3, H1, H2}-free, (G) = 2, and (G) = 3 = (G)+1. Examples to show that K1,3, H1, H2 are Necessary:  Mycielski graphs with (G) = k  4, have only K1,3-induced, are triangle-free, hence have (G) = 2 < (G) - 1. 

Let H = C5. Construct G from H by replacing each vertex by Km. Then G is {K1,3, H2}-free, G has H1 induced, (G) = 2m, and (G) = (G)+



m

Let G =

C

5

m  > (G)+1 for m  3. 2

. Then G is {K1,3, H1}-free, G has H2 induced, (G) = 2m, and (G) = 3m >

1

(G)+1 for m > 1. References [1]. “Chromatic bounds for a class of graphs”, S. A. Choudum, Quart Jour. Math. Oxford Ser. 28, 3 (1977), 257-270. [2]. “A note on Choudum’s chromatic bounds for a class of graphs”, M. Javdekar, Jour. of Graph Theory 4 (1980), 265-268. [3]. “On the chromatic index of multigraphs without large triangles”, H. Kierstead, Jour. of Comb. Theory Ser. B 36 (1984), 156-160. [4]. “On an estimate of the chromatic number of a p-graph”, V. G. Vizing, Discrete Anal. 3 (1964), 2530. [5] “On the Chromatic Number of a Graph with Two Forbidden Subgraphs”, Medha Dhurandhar, Jour. of Comb. Theory, Ser. B 46 (1989), 1-6.